- #1

Eclair_de_XII

- 1,083

- 91

- TL;DR Summary
- Define a function ##f:D\longrightarrow \mathbb{R}## such that for some ##a\in D##, ##f(x)>0## if ##x>a## and ##f(x)\leq 0## if ##x\leq a##. Note: ##f(a)=0## by the IVT. If ##|f|## is uniformly continuous on ##D##, then does that mean ##f## is also uniformly continuous?

I'd say yes, it is. Suppose ##|f|## is uniformly continuous on ##D##.

Then for all ##\epsilon>0## there is ##\delta>0## (call this ##\delta'##) such that if ##x,y\in D##, then ##||f(x)|-|f(y)||<\epsilon##.

Define sets:

##D^+=\{x\in D: x>a\}##

##D^-=\{x\in D: x<a\}##

Restrict the domain of ##f## to ##D^+##; note that ##f=|f|## here so it follows that ##f## is uniformly continuous here. In other words, if you want the distance between the images any two points ##x,y\in D^+## to be as small as possible, you need only ensure that the distance between ##x,y## is at most ##\delta'##.

Now restrict the domain of ##f## to ##D^-##. Let ##\epsilon>0## and choose ##\delta'##. Note that if ##x,y\in D^-##, then:

\begin{align}

|f(x)-f(y)|&=&|-|f(x)|-(-|f(y))|\\

&=&|-|f(x)|+|f(y)|| \\

&<&\epsilon

\end{align}

since ##|f|## is uniformly continuous.

Finally, note that ##|f|## is continuous at ##x=a## and moreover, to ensure that the image of a point ##x## is within ##\epsilon## of ##f(a)=0##, you can choose an ##x## within ##(a-\delta',a+\delta')##, since ##|f|## is uniformly continuous.

Let ##\epsilon>0## and choose ##\delta=\delta'>0##. Suppose ##x\in D^+ \cup D^-##. Then:

\begin{align}

|f(x)-f(a)|&=&|f(x)|\\

&=&||f|(x)|\\

&<&\epsilon

\end{align}

Then for all ##\epsilon>0## there is ##\delta>0## (call this ##\delta'##) such that if ##x,y\in D##, then ##||f(x)|-|f(y)||<\epsilon##.

Define sets:

##D^+=\{x\in D: x>a\}##

##D^-=\{x\in D: x<a\}##

Restrict the domain of ##f## to ##D^+##; note that ##f=|f|## here so it follows that ##f## is uniformly continuous here. In other words, if you want the distance between the images any two points ##x,y\in D^+## to be as small as possible, you need only ensure that the distance between ##x,y## is at most ##\delta'##.

Now restrict the domain of ##f## to ##D^-##. Let ##\epsilon>0## and choose ##\delta'##. Note that if ##x,y\in D^-##, then:

\begin{align}

|f(x)-f(y)|&=&|-|f(x)|-(-|f(y))|\\

&=&|-|f(x)|+|f(y)|| \\

&<&\epsilon

\end{align}

since ##|f|## is uniformly continuous.

Finally, note that ##|f|## is continuous at ##x=a## and moreover, to ensure that the image of a point ##x## is within ##\epsilon## of ##f(a)=0##, you can choose an ##x## within ##(a-\delta',a+\delta')##, since ##|f|## is uniformly continuous.

Let ##\epsilon>0## and choose ##\delta=\delta'>0##. Suppose ##x\in D^+ \cup D^-##. Then:

\begin{align}

|f(x)-f(a)|&=&|f(x)|\\

&=&||f|(x)|\\

&<&\epsilon

\end{align}